Question: $f(x)=\begin{cases} x^2&\text{for }x\leq0 \\\\ \text{ln}(x)&\text{for }x>0 \end{cases}$ Find $\lim_{x\to 1}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $e$ (Choice D) D The limit doesn't exist.
Solution: Let's find the limit as $x$ approaches $1$. We will use the fact that $f(x)=\text{ln}(x)$ for $x$ -values larger than $0$. $\begin{aligned} &\phantom{=}\lim_{x\to 1}f(x) \\\\ &=\lim_{x\to 1}\text{ln}(x) \\\\ &=\text{ln}(1)&\gray{\text{Direct substitution}} \\\\ &=0 \end{aligned}$ In conclusion, we found that $\lim_{x\to 1}f(x)=0$.